Question
.
Let $z_1 = 4 + 5i$.
(a) (i) Find $|z_1|$.
(ii) Find arg(z_1).
Let $z_2 = 3e^{\frac{2\pi}{3}i}$.
(b) Find the area of the triangle on an Argand diagram with vertices 0, $z_1$ and $z_2$.
▶️Answer/Explanation
Detailed Solution
Part (a): Analyzing \( z_1 = 4 + 5i \)
(i) Find \( |z_1| \)
The modulus (or magnitude) of a complex number \( z = a + bi \) is given by:
\[ |z| = \sqrt{a^2 + b^2} \]
For \( z_1 = 4 + 5i \):
\( a = 4 \),
\( b = 5 \).
\[ |z_1| = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \]
So, the modulus is:
\[ |z_1| = \sqrt{41} \]
(ii) Find \( \arg(z_1) \)
The argument of a complex number \( z = a + bi \) is the angle \( \theta \) that the line from the origin to the point \((a, b)\) makes with the positive real axis, given by:
\[ \tan \theta = \frac{b}{a} \]
For \( z_1 = 4 + 5i \):
\( a = 4 \),
\( b = 5 \).
\[ \tan \theta = \frac{5}{4} \]
\[ \theta = \tan^{-1}\left(\frac{5}{4}\right) \]
Since \( a = 4 > 0 \) and \( b = 5 > 0 \), \( z_1 \) lies in the first quadrant, so \( \theta \) is between 0 and \( \frac{\pi}{2} \). We leave it as \( \tan^{-1}\left(\frac{5}{4}\right) \)
\[ \arg(z_1) = \tan^{-1}\left(\frac{5}{4}\right) \]
Part (b): Area of the Triangle with Vertices 0, \( z_1 \), and \( z_2 \)
We need to find the area of the triangle on the Argand diagram with vertices at:
The origin: \( 0 = 0 + 0i \),
\( z_1 = 4 + 5i \),
\( z_2 = 3e^{\frac{2\pi}{3}i} \).
Step 1: Convert \( z_2 \) to Rectangular Form
The complex number \( z_2 = 3e^{\frac{2\pi}{3}i} \) is in polar form, where:
Magnitude = 3,
Argument = \( \frac{2\pi}{3} \).
Using Euler’s formula, \( e^{i\theta} = \cos \theta + i \sin \theta \):
\[ z_2 = 3 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) \]
Compute the trigonometric values:
\( \cos \frac{2\pi}{3} = \cos 120^\circ = -\frac{1}{2} \),
\( \sin \frac{2\pi}{3} = \sin 120^\circ = \frac{\sqrt{3}}{2} \).
So:
\[ z_2 = 3 \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = -\frac{3}{2} + i \frac{3\sqrt{3}}{2} \]
Step 2: Assign Coordinates
On the Argand diagram, complex numbers correspond to points in the plane (real part as \( x \), imaginary part as \( y \)):
Origin: \( 0 = (0, 0) \),
\( z_1 = 4 + 5i = (4, 5) \),
\( z_2 = -\frac{3}{2} + i \frac{3\sqrt{3}}{2} = \left( -\frac{3}{2}, \frac{3\sqrt{3}}{2} \right) \).
Step 3: Compute the Area of the Triangle
The area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be found using the determinant formula:
\[ \text{Area} = \frac{1}{2} \left| x_1 (y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2) \right| \]
Assign:
\( (x_1, y_1) = (0, 0) \) (origin),
\( (x_2, y_2) = (4, 5) \) (\( z_1 \)),
\( (x_3, y_3) = \left( -\frac{3}{2}, \frac{3\sqrt{3}}{2} \right) \) (\( z_2 \)).
Substitute into the formula:
\[ \text{Area} = \frac{1}{2} \left| 0 \cdot \left( 5 – \frac{3\sqrt{3}}{2} \right) + 4 \cdot \left( \frac{3\sqrt{3}}{2} – 0 \right) + \left( -\frac{3}{2} \right) \cdot (0 – 5) \right| \]
Simplify each term:
– First term: \( 0 \cdot \left( 5 – \frac{3\sqrt{3}}{2} \right) = 0 \),
– Second term: \( 4 \cdot \left( \frac{3\sqrt{3}}{2} – 0 \right) = 4 \cdot \frac{3\sqrt{3}}{2} = \frac{12\sqrt{3}}{2} = 6\sqrt{3} \),
– Third term: \( -\frac{3}{2} \cdot (0 – 5) = -\frac{3}{2} \cdot (-5) = \frac{15}{2} \).
Combine:
\[ \text{Area} = \frac{1}{2} \left| 0 + 6\sqrt{3} + \frac{15}{2} \right| = \frac{1}{2} \left| 6\sqrt{3} + \frac{15}{2} \right| \]
Since the expression is positive (both terms are positive), the absolute value is unnecessary:
\[ 6\sqrt{3} + \frac{15}{2} \]
\[ \text{Area} = \frac{1}{2} \left( 6\sqrt{3} + \frac{15}{2} \right) = \frac{6\sqrt{3}}{2} + \frac{15}{4} = 3\sqrt{3} + \frac{15}{4} \] = $8.58 $
………………………..Markscheme……………………………
Solution: –
$(a) \ (i) \ |z_1| = \sqrt{4^2 + 5^2} = 6.40 (6.40312…, \sqrt{41})$
$(ii) \ arg(z_1) = 0.896 (0.896055…, 51.3401…, arctan(\frac{5}{4}))$
$(b) \ \text{angle in triangle is } 2-0.896055… \text{ OR } 114.591-51.3401…$
$\text{use of area of triangle formula}$
$\frac{1}{2} \times 6.40312… \times 3 \times \sin(2-0.896055…)$
$8.58 (8.57688…)$
Question
Consider the complex number \( z = -1 + i \).
(a) Express \( z \) in the form \( r e^{i\theta} \) where \( -\pi < \theta \leq \pi \).
A and B are the points on the Argand diagram that represent the complex numbers \( z \) and \( z^2 \), respectively.
A is mapped onto B by the composition of a rotation and an enlargement.
(b) (i) Describe fully this mapping of A onto B, stating the scale factor of the enlargement and the angle of rotation.
(ii) Find and simplify a matrix that maps A onto B.
(c) Find the smallest positive integer, \( n \), for which \( z^n \) is real and positive.
▶️ Answer/Explanation
Detailed Solution
(a) Expressing \( z \) in polar form
- Magnitude: \( r = \sqrt{(-1)^2 + (1)^2} = \sqrt{2} \).
- Argument: \( \theta = \tan^{-1} \left(\frac{1}{-1}\right) = \frac{3\pi}{4} \).
Thus, \( z \) can be written as:
\[ z = \sqrt{2} e^{\frac{3\pi i}{4}} \]
(b) (i) Mapping A onto B
- Enlargement: Scale factor \( \sqrt{2} \) (approximately 1.41421), centered at (0,0).
- Rotation: \( \frac{3\pi}{4} \) radians (approximately 2.35619), counterclockwise about the origin.
(ii) Finding the transformation matrix
The transformation consists of an enlargement by \( \sqrt{2} \) followed by a rotation by \( \frac{3\pi}{4} \), represented by the matrices:
\[ \begin{pmatrix} \sqrt{2} & 0 \\ 0 & \sqrt{2} \end{pmatrix} \] \[ \begin{pmatrix} \cos \frac{3\pi}{4} & -\sin \frac{3\pi}{4} \\ \sin \frac{3\pi}{4} & \cos \frac{3\pi}{4} \end{pmatrix} \]
(c) Finding the smallest \( n \) for which \( z^n \) is real and positive
- \( z^n = (\sqrt{2} e^{\frac{3\pi i}{4}})^n = (\sqrt{2})^n e^{\frac{3n\pi i}{4}} \).
- For \( z^n \) to be real and positive, the argument must be a multiple of \( 2\pi \):
\[ \frac{3n\pi}{4} = 2k\pi, \quad k \in \mathbb{Z} \]
- Solving for \( n \):
\[ n = \frac{8k}{3}, \quad k = 3 \Rightarrow n = 8 \]
Final Answer: The smallest positive integer \( n \) is 8.
…………………………..Markscheme…………………………..
(a)
- Correct calculation of magnitude \( \sqrt{2} \).
- Correct argument \( \frac{3\pi}{4} \).
- Final answer: \( \sqrt{2} e^{\frac{3\pi i}{4}} \).
(b) (i)
- Correct scale factor: \( \sqrt{2} \).
- Correct rotation: \( \frac{3\pi}{4} \) radians.
(b) (ii)
- Correct transformation matrix.
(c)
- Correct derivation of the argument condition.
- Correct integer value: \( n = 8 \).